Problem: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x - 4)^2 - 25$ $\text{lesser }x = $
Answer: $\begin{aligned} (x - 4)^2 - 25&= 0 \\\\ (x-4)^2&=25 \\\\ \sqrt{(x-4)^2}&=\sqrt{25} \end{aligned}$ $\begin{aligned} x-4&=\pm5 \\\\ x&=\pm5+4 \\ \phantom{(x - 4)^2 - 25}& \\ x=-1&\text{ or }x=9 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -1 \\\\ \text{greater } x &= 9 \end{aligned}$